$S$ and $T$ are points on sides $PR$ and $QR$ of $\Delta PQR$ such that $\angle P = \angle RTS$. Show that $\Delta RPQ \sim \Delta RTS$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then the correspondence $ADB \leftrightarrow \ldots$ between $\Delta ADB$ and $\Delta BDC$ is a similarity.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow QPR.$ If $m \angle P : m \angle Q : m \angle R = 2 : 3 : 4,$ then $m \angle A : m \angle B : m \angle C = \ldots$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 25$ and $AM = 16$,then $BM = \dots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$ and $BM$.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. $\overline{AD}$ and $\overline{PM}$ are medians of these triangles. Prove that $AB \times PM = PQ \times AD$.